📚 A-Level Maths: Gradients, Tangents and Normals | A-Level数学：梯度、切线与法线问题
In coordinate geometry and calculus, the gradient of a curve at a point measures how steeply the curve is rising or falling there. For a curve, the gradient is not constant, so we use differentiation to find the gradient at a specific point. Once the gradient is known, we can write down the equation of the tangent, and from it the equation of the normal, which is perpendicular to the tangent.
在坐标几何和微积分中，曲线在某一点处的梯度衡量曲线在该点上升或下降的陡峭程度。曲线的梯度并不是恒定的，因此我们需要通过微分求某一点的梯度。在求出梯度后，我们可以写出切线的方程，并由此得到法线的方程，因为法线与切线互相垂直。
1. The Gradient at a Point | 曲线在某点的梯度
For a straight line, the gradient is constant and can be found by dividing the change in y by the change in x. For a curve, the ratio of change in y to change in x over an interval gives only an average gradient, not the exact gradient at a single point.
对于直线，梯度是恒定的，可以用 y 的变化量除以 x 的变化量来求得。对于曲线，在一段区间内 y 的变化量与 x 的变化量之比只能给出平均梯度，而不是某一点处的精确梯度。
The exact gradient of the curve y = f(x) at x = a is the limit of the average gradient as the interval length tends to zero. This is written as:
曲线 y = f(x) 在 x = a 处的精确梯度是区间长度趋于零时平均梯度的极限，写为：
f'(a) = lim[h→0] ( f(a+h) − f(a) ) / h
If this limit exists, the derivative f'(a) gives the gradient of the curve at that point. It is also the slope of the tangent line drawn through the point.
如果这个极限存在，那么导数 f'(a) 就给出曲线在该点的梯度，也就是经过该点的切线的斜率。
2. Finding the Gradient by Differentiation | 通过微分求梯度
In A-Level Mathematics, the most efficient way to find the gradient is to differentiate the function. The key rule is the power rule: if y = xⁿ, then dy/dx = n xⁿ⁻¹.
在 A-Level 数学中，求梯度的最有效方法是对函数求导。核心法则是幂函数法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。
This rule also works for constants, negative powers and fractional powers. Some common derivatives are shown below.
这个法则同样适用于常数、负指数和分数指数。下面是一些常见函数的导数。
| y | dy/dx |
| x² | 2x |
| x³ | 3x² |
| x⁻² | −2x⁻³ |
| √x = x^(1/2) | 1/(2√x) = ½ x^(−1/2) |
| 5 | 0 |
| x^(3/2) | (3/2)x^(1/2) |
To differentiate a sum of terms, differentiate each term separately. If y = f(x) + g(x), then dy/dx = f'(x) + g'(x). If y = k f(x), then dy/dx = k f'(x), where k is a constant.
对多项式求和求导时，逐项求导即可。若 y = f(x) + g(x)，则 dy/dx = f'(x) + g'(x)。若 y = k f(x)，则 dy/dx = k f'(x)，其中 k 是常数。
3. Equation of a Tangent | 切线方程
A tangent is a straight line that touches a curve at exactly one point and has the same gradient as the curve at that point. To find its equation, you need the gradient at the point and the coordinates of the point.
切线是一条直线，它与曲线恰好接触于一点，并且在该点处与曲线具有相同的梯度。要求切线方程，你需要知道该点处曲线的梯度以及该点的坐标。
Suppose the point is (x₁, y₁) and the gradient at that point is m. The equation of the tangent is given by the point-slope form:
设点为 (x₁, y₁)，该点处的梯度为 m，则切线方程为点斜式：
y − y₁ = m(x − x₁)
You may need to rearrange this into the form y = ax + b. When finding the tangent, first substitute the given x-coordinate into the original function to find y₁, then substitute the same x-coordinate into dy/dx to find m.
你可能需要把它重新整理成 y = ax + b 的形式。求切线时，先把给定的 x 坐标代入原函数求 y₁，再把同一个 x 坐标代入 dy/dx 求 m。
4. Equation of a Normal | 法线方程
The normal at a point on a curve is the straight line perpendicular to the tangent at that point. If the tangent gradient is m, then the normal gradient is the negative reciprocal of m.
曲线上某一点处的法线，是经过该点并与切线垂直的直线。如果切线梯度为 m，那么法线梯度是 m 的负倒数。
If m₁ × m₂ = −1, then the two straight lines are perpendicular.
因此，如果切线梯度为 m，则法线梯度为 −1/m，法线方程为：
y − y₁ = −1/m (x − x₁)
This formula is only valid when m ≠ 0. If m = 0, the tangent is horizontal and the normal is the vertical line x = x₁.
这个公式只在 m ≠ 0 时成立。如果 m = 0，切线是水平线，法线就是竖直线 x = x₁。
5. Worked Example: Tangent and Normal to y = x³ − 2x + 3 | 完整例题：求 y = x³ − 2x + 3 的切线与法线
Find the equations of the tangent and the normal to the curve y = x³ − 2x + 3 at the point where x = 1.
求曲线 y = x³ − 2x + 3 在 x = 1 处的切线方程和法线方程。
Step 1: Find the y-coordinate.
第一步：求 y 坐标。
y = 1³ − 2(1) + 3 = 1 − 2 + 3 = 2, so the point is (1, 2).
y = 1³ − 2(1) + 3 = 1 − 2 + 3 = 2，因此点为 (1, 2)。
Step 2: Differentiate to find the gradient.
第二步：求导得到梯度。
dy/dx = 3x² − 2. At x = 1: m = 3(1)² − 2 = 3 − 2 = 1.
dy/dx = 3x² − 2。在 x = 1 处：m = 3(1)² − 2 = 3 − 2 = 1。
Step 3: Write down the tangent equation.
第三步：写出切线方程。
y − 2 = 1(x − 1), so y = x + 1.
y − 2 = 1(x − 1)，所以 y = x + 1。
Step 4: Use the negative reciprocal for the normal.
第四步：用负倒数求法线。
The gradient of the normal is −1. Therefore y − 2 = −1(x − 1), which simplifies to y = −x + 3.
法线梯度为 −1。因此 y − 2 = −1(x − 1)，化简得 y = −x + 3。
Check: the two gradients 1 and −1 have product −1, so the tangent and normal are perpendicular.
检验：两个梯度 1 和 −1 的乘积等于 −1，因此切线与法线互相垂直。
6. Special Cases: Horizontal and Vertical Tangent or Normal | 特殊情况：水平与竖直的切线或法线
If dy/dx = 0 at a point, the tangent is horizontal. The normal is then vertical and has the equation x = x₁. This often happens at a stationary point, such as a maximum, minimum or point of inflection.
如果在某点处 dy/dx = 0，则切线是水平的。此时法线是竖直的，方程为 x = x₁。这种情况常出现在驻点，例如极大值点、极小值点或拐点。
If dy/dx is undefined at a point, the tangent may be vertical and the normal horizontal. For example, for y = x^(1/3) at x = 0, dy/dx = (1/3)x^(−2/3), which tends to infinity as x → 0. The tangent is x = 0 and the normal is y = 0.
如果在某点处 dy/dx 不存在，切线可能是竖直的，而法线是水平的。例如，对于 y = x^(1/3)，在 x = 0 处，dy/dx = (1/3)x^(−2/3)，当 x → 0 时该值趋于无穷大。因此切线为 x = 0，法线为 y = 0。
When you use the normal formula, always check whether m = 0 before writing −1/m. Division by zero is not defined, so you must use the vertical or horizontal line form instead.
在使用法线公式时，请先检查 m 是否等于 0，再写出 −1/m。除以零没有意义，因此这种情况下必须改用竖直线或水平线方程。
7. Common Pitfalls and How to Avoid Them | 常见错误与避免方法
Tangent and normal problems appear frequently in Edexcel A-Level exams. Students often lose marks not because the differentiation is hard, but because of small mistakes in substitution or sign.
切线与法线问题在 Edexcel A-Level 考试中频繁出现。学生丢分往往不是因为求导本身有多难，而是因为代入或符号上的小错误。
| Common mistake 常见错误 | Correction 正确做法 |
| Substituting x into the original equation to find the gradient. | Substitute x into dy/dx, not into y. |
| Forgetting to find the y-coordinate of the point. | Substitute x into the original curve equation to get y₁. |
| Using 1/m for the normal gradient instead of −1/m. | The product of tangent and normal gradients must be −1. |
| Writing m = 0 and then trying to calculate −1/0. | If m = 0, the normal is vertical: x = x₁. |
| Using the wrong coordinates when substituting into the straight-line equation. | Use
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